A definite integral $$\int_0^1\sqrt{\left(3-3t^2\right)^2+\left(6t\right)^2}\,dt$$ I am trying to take this integral. I know the answer is 4. But I am having trouble taking the integral itself. I've tried foiling and the simplifying. I've tried u-sub. I just can't get the correct way to take the integral. Any help would be appreciated. Sorry if the layout doesn't look right.

Hint : By doing some manipulation and expansion, notice that $$\begin{align} (3-3t^2)^2 + (6t)^2 &= 3^2(1-2t^2 +t^4 +4t^2)\\\ &= 9(1+ 2t^2 + t^4) \\\ &=9(1+t^2)^2\end{align}$$

Now, just take the square root of this and integrate the result.